Question

Cho \(A=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\). Chứng minh 50 < A < 100

Answer 1

Ta có:

A = 1 + ( 1/2 + 1/3 ) + ( 1/4 + 1/5 + 1/6 + 1/7 ) + ( 1/8 + 1/9 +... + 1/15 )+...+ ( 1/2^99 + 1/2^99 + 1 +...+ 1/2^100 - 1 )
( Có 99 nhóm ) < 1+2.1/2+2^2.1/2^2+2^3.1/2^3+.....+2^99.1/2^99
=> 1 + 1 + 1 + ... + 1 (100 số 1) = 100
=> A1 + 1/2 + 2.1/2^2+2^2.1/2^3 + 2^3 . 1/2^4 +... + 2^991/2^100 - 1 -1/2^100

= 1 + 1/2 + 1/2 + 1/2 + 1/2 + ... + 1/2 - 1/2^100 (100 số 1/2)
= 1 + 100 . 12 - 1/2^100
= 50 + 1 - 1/2^100 > 50
=> A > 50 (2)
Từ (1) và (2) => 50